Chemical kinetics, order confusion
In the solution they've just directly used the first order equation (since overall order = 1) and got the answer. How did that work? Isn't the reaction half order with respect to A, so it should behave as such?
31 Replies
@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.order wrt A has no meaning , its just order
wot
Order wrt A is like a Question
ive seen that terminology used in questions
it has to have some meaning
i meant
it does not matter
the overall order of reaction dictates the overall rate
evidently not yeah
but i dont conceptually get why
ic what u are trying to say
my thought process was that if A is being consumed at a half order rate and so is B then to get details from A, we have to take its consumption behaviour into account
have u tried using that
uh rate of formation = rate of being used
hain?
A aint being formed only being used
ye ik ,
voh toh consecutive rxn mein hota hai na
wait tht wont work
lemme try smth
ngl it does feel counterintuivtive when we talk about A only
yeh
i tried dA/dt = K[A]^1/2 and then integration with initial and final concentration limits
but i mean that just gave negative time and magnitude is also not the answer
what is the ans
But dA/dt=k[A]^0.5 [B]^0.5
Which means [A] depends on [B] too
In this case [A]=[B]
Here dB/dt=dA/dt so, A=B+c
From initial condition, c=0, so the rate of reaction depends on k[A]^0.5[B]^0.5=k[A] (since [A]=[B])
If you change the initial concentration ki condition, answer would be different.
ahhh
oke yeah that makes perfect sense
gotcha, thank u
50sec
@Gamertug explanation above ^ lmk if u understood, ill close it then
ok correct aya
mast
they're being consumed in an identical manner so its basically just equivalent to k[A] anyway
so what if c wasnt equal to 0
suppose if B started out at 2 molar instead of 1
Yeah but beware of situations where initial concentrations are not same. Then that would be done by differential rate law.
reference point hai i think
Then dA/dt=ksqrt(A)sqrt(A+1) and then you solve to get dA/sqrt((A+1/2)^2-1/4)=kdt which gives you ln(A+sqrt(A^2+A)) with limits=k*t.
that makes sense
na actually wait
arent there any number of functions between 2 numbers
1 to 2 is a simple example but there are prob other alternatives compared to A+1
This point?
yeah nvm
got it
Ok but what if it’s not like just A+1 but more like to the point you ignore it
thx
ahh ic
+solved SirLancelotDuLac
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